证明:如图,连结AC1交A1M于P,
∵∠BCA=90°,且BC⊥CC1,
∴BC⊥平面AA1C1C.
又BC∥B1C1,从而有B1C1⊥平面AA1C1C.
∴AC1是AB1在平面AA1C1C上的射影.
设∠AC1A1=α,∠MA1C1=β,则tanα=,tanβ= .
∴α+β=90°,即AC1⊥A1M.
∴AB1⊥A1M.
,M为CC1的中点,求证:AB1⊥A
,M为CC1的中点,求证:AB1⊥A,tanβ=
.